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Chapter 13: Problem 38

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes as follows to nitrogen dioxide and nitrogen trioxide: $$ \mathrm{N}_{2} \mathrm{O}_{5}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g) $$ Calculate the average rate of this reaction between consecutive measurement times in the following table. $$\begin{array}{cc} \text { Time (s) } & {\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]} \\ & \text { (molecules/cm }^{3} \text { ) } \\ 0.00 & 1.500 \times 10^{12} \\ \hline 1.45 & 1.357 \times 10^{12} \\ \hline 2.90 & 1.228 \times 10^{12} \\ \hline 4.35 & 1.111 \times 10^{12} \\ \hline 5.80 & 1.005 \times 10^{12} \\ \hline \end{array}$$

Short Answer

Expert verified
Question: Calculate the average rate of reaction for each consecutive time interval for the decomposition of dinitrogen pentoxide based on the given concentration values at different times. Answer: The average rate of reaction for each consecutive time interval is as follows: - Interval 1: \(-0.099\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 2: \(-0.089\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 3: \(-0.081\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 4: \(-0.073\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\)

Step by step solution

01

Identify the consecutive time intervals and corresponding concentration values

In the table given, there are consecutive measurement times and their corresponding concentrations. We need to find the average rate of reaction for each interval: - Interval 1: 0.00s to 1.45s - Interval 2: 1.45s to 2.90s - Interval 3: 2.90s to 4.35s - Interval 4: 4.35s to 5.80s
02

Calculate the average rate of reaction for each time interval

Use the formula: \(\text{average rate} = \frac{\Delta \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]}{\Delta t}\), where \(\Delta \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) is the change in concentration during the time interval and \(\Delta t\) is the change in time during the interval. - Interval 1: Average rate = \(\frac{(1.357\times 10^{12})-(1.500\times 10^{12})}{(1.45-0)}\) = \(\frac{-0.143\times 10^{12}}{1.45}\) = \(-0.099\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 2: Average rate = \(\frac{(1.228\times 10^{12})-(1.357\times 10^{12})}{(2.9-1.45)}\) = \(\frac{-0.129\times 10^{12}}{1.45}\) = \(-0.089\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 3: Average rate = \(\frac{(1.111\times 10^{12})-(1.228\times 10^{12})}{(4.35-2.90)}\) = \(\frac{-0.117\times 10^{12}}{1.45}\) = \(-0.081\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 4: Average rate = \(\frac{(1.005\times 10^{12})-(1.111\times 10^{12})}{(5.80-4.35)}\) = \(\frac{-0.106\times 10^{12}}{1.45}\) = \(-0.073\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\)
03

Present the results with the average rate of reaction for each time interval

The average rate of reaction for each consecutive time interval is as follows: - Interval 1: \(-0.099\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 2: \(-0.089\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 3: \(-0.081\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\) - Interval 4: \(-0.073\times 10^{12}\ \text{molecules/cm}^3\text{s}^{-1}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Kinetics
Chemical kinetics is the study of the speed or rate at which chemical reactions occur, and the factors that affect these rates. It's a pivotal field in chemistry because understanding how fast reactions proceed helps us to design and control chemical processes, from the metabolism of substances in our body to industrial synthesis of chemicals.

When studying chemical kinetics, we often focus on reaction rates, the changes in concentrations of reactants or products over time, and how factors such as temperature, pressure, and concentration influence these rates. By experimenting and building mathematical models, chemists can predict reaction behavior and optimize conditions for desired outcomes.
Rate of Reaction
The rate of a chemical reaction indicates the speed at which reactants are converted into products. It's generally expressed as a change in concentration of a reactant or product per unit time. When we talk about an average rate, we're referring to the amount of reactant used or product formed over a longer time interval. Mathematically, this is often presented as the negative change in concentration of the reactant divided by the change in time, due to the reactant concentration decreasing over time.

In chemical kinetics, it's crucial to understand that the rate of reaction can vary as the conditions or concentrations of substances change during the reaction. This is why rates are specified at particular moments or intervals.
Concentration Changes Over Time
In kinetics, changes in concentration over time are central to understanding how reactions progress. Concentration typically decreases for reactants and increases for products. By measuring these concentration changes at regular intervals, we can track the reaction's progress and calculate the reaction rates.

To calculate these changes, the concentration at the start and end of a time interval is determined, and the difference is divided by the duration of the interval. Monitoring these changes helps chemists adjust conditions to control the reaction rate, which is especially important in industrial applications to ensure safety and efficiency.
Dinitrogen Pentoxide Decomposition
The decomposition of dinitrogen pentoxide (\(\mathrm{N}_2\mathrm{O}_5\)) is a chemical process where this compound breaks down into nitrogen dioxide (\(\mathrm{NO}_2\)) and a nitrogen trioxide species. This reaction has been well-studied in kinetics because of its relatively simple stoichiometry and its occurrence in atmospheric chemistry.

Calculating the reaction rate for dinitrogen pentoxide decomposition involves measuring the concentration of \(\mathrm{N}_2\mathrm{O}_5\) at different times and using these values to determine the rate at which it's decreasing. This is a classic example of a first-order reaction, where the rate is directly proportional to the concentration of the reactant. In educational settings, this decomposition serves as an excellent model for teaching fundamental kinetics concepts.

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Most popular questions from this chapter

During a reaction, can there be a time when the instantaneous rate of the reaction does not change? If you think so, describe such a time.

Sources of Nitric Acid in the Atmosphere The reaction between \(\mathrm{N}_{2} \mathrm{O}_{5}\) and water, $$ \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightarrow 2 \mathrm{HNO}_{3}(g) $$ is a source of nitric acid in the atmosphere. a. The reaction is first order in each reactant. Write the rate law for the reaction. b. When \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\) is \(0.132 \mathrm{m} \mathrm{M}\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]\) is \(230 \mathrm{m} M,\) the rate of the reaction is \(4.55 \times 10^{-4} \mathrm{m} M^{-1} \mathrm{min}^{-1} .\) What is the rate constant for the reaction?

Compounds \(A\) and \(B\) react to give a single product, \(C .\) Write the rate law for each of the following cases and determine the units of the rate constant by using the units \(M\) for concentration and s for time: a. The reaction is first order in \(A\) and second order in \(B\). b. The reaction is first order in \(A\) and second order overall. c. The reaction is independent of the concentration of \(\mathrm{A}\) and second order overall. d. The reaction is second order in both \(\mathrm{A}\) and \(\mathrm{B}\).

In a study of the thermal decomposition of ammonia into nitrogen and hydrogen: $$ 2 \mathrm{NH}_{3}(g) \rightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) $$ the average rate of change in the concentration of ammonia is \(-0.38 M / s.\) a. What is the average rate of change in \(\left[\mathrm{H}_{2}\right] ?\) b. What is the average rate of change in \(\left[\mathrm{N}_{2}\right] ?\) c. What is the average rate of the reaction?

Nitrogen Oxide in the Human Body Nitrogen oxide is a free radical that plays many biological roles, including regulating neurotransmission and the human immune system. One of its many reactions involves the peroxynitrite ion (ONOO"): $$ \mathrm{NO}(a q)+\mathrm{ONOO}^{-}(a q) \rightarrow \mathrm{NO}_{2}(a q)+\mathrm{NO}_{2}^{-}(a q) $$ a. Use the following data to determine the rate law and rate constant of the reaction at the experimental temperature at which these data were generated. $$\begin{array}{cccc} \hline \text { Experiment } & \text { [NOlo (M) } & \text { [ONOO Jo (M) } & \text { Rate (M/s) } \\ \hline 1 & 1.25 \times 10^{-4} & 1.25 \times 10^{-4} & 2.03 \times 10^{-11} \\\ \hline 2 & 1.25 \times 10^{-4} & 0.625 \times 10^{-4} & 1.02 \times 10^{-11} \\\ \hline 3 & 0.625 \times 10^{-4} & 2.50 \times 10^{-4} & 2.03 \times 10^{-11} \\\ \hline 4 & 0.625 \times 10^{-4} & 3.75 \times 10^{-4} & 3.05 \times 10^{-11} \\\ \hline \end{array}$$ b. Draw the Lewis structure of peroxynitrite ion (including all resonance forms) and assign formal charges. Note which form is preferred. c. Use the average bond energies in Table A4.1 to estimate the value of \(\Delta H_{\text {rxn }}\) by using the preferred structure from part (b).
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